Let $S^1 \times S^1$ be the $2$-torus. If a point $a=(p,q)$ of the torus is removed, i.e., it is punctured at one point then how can I show that it can be immersed in the plane, i.e., in $\Bbb{R}^2$?

I have the idea of decomposing the punctured torus into two intersecting cylinders and then immersing each of them to $\Bbb{R}^2$, but the problem is by this way even if we make that the intersecting region mapped to same region, we cannot define an immersion on the whole the two immersions may not agree pointwise on the intersection. I have got stuck here.

1 Answer
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What follows is a topologist's solution, but I believe quite revealing.

A torus minus one point is a torus minus a disc. Expanding the disc one gets two cilinders (without border, but for the picture it's easier to draw it) attached as seen. These are just a couple of annuli with a common rectangle, which can be projected onto the plane. It's an open inmersion, of course not injective.